Let a random sample of n 100 be drawn from a population with

Let a random sample of n 100 be drawn from a population with mean 50 and variance 1. If X is the sample mean, calculate approximately P(49.75l e X le 50.25). The probability of a computer chip made in a certain plant will be defective is 0.05. If a sample of 1,000 chips is tested, obtain approximate probability that less than 40 will be defective by using the normal approximation to the Binomial distribution.

Solution

5.

A)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    49.75      
x2 = upper bound =    50.25      
u = mean =    50      
n = sample size =    100      
s = standard deviation =    1      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.5      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.006209665      
P(z < z2) =    0.993790335      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.987580669   [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    39.5      
u = mean = np =    50      
          
s = standard deviation = sqrt(np(1-p)) =    6.892024376      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.523500125      
          
Thus, the left tailed area is          
          
P(z <   -1.523500125   ) =    0.063816815 [ANSWER]